Корень(4+корень7)-корень(4-корень7)

To simplify the expression $\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}$ , we can use the method of squaring the entire expression or transforming the nested radicals into perfect squares. Method 1: Squaring the Expression Let the given expression be $x$: $x=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}$ Now, square both sides of the equation using the formula $(a-b{)}^2=a^2-2ab+b^2$: $x^2=(\sqrt{4+\sqrt{7}}{)}^2-2(\sqrt{4+\sqrt{7}}\left)\right(\sqrt{4-\sqrt{7}})+(\sqrt{4-\sqrt{7}}{)}^2$

1. Simplify the squared terms:
   $(\sqrt{4+\sqrt{7}}{)}^2=4+\sqrt{7}$ $(\sqrt{4-\sqrt{7}}{)}^2=4-\sqrt{7}$ Simplify the middle term (product of radicals):
   Using the property $\sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b}$ and the difference of squares $(a-b)(a+b)=a^2-b^2$:
   $2\sqrt{(4+\sqrt{7})(4-\sqrt{7})}=2\sqrt{4^2-(\sqrt{7}{)}^2}=2\sqrt{16-7}=2\sqrt{9}$ $2\sqrt{9}=2\cdot 3=6$ Combine all parts:
   $x^2=(4+\sqrt{7})-6+(4-\sqrt{7})$ $x^2=4+\sqrt{7}-6+4-\sqrt{7}$ $x^2=4-6+4=2$ Solve for $x$:
   Since $\sqrt{4+\sqrt{7}}>\sqrt{4-\sqrt{7}}$ , the result must be positive:
   $x=\sqrt{2}$

Method 2: Transforming Nested Radicals We can rewrite the expressions inside the square roots to create perfect squares. To do this, we multiply and divide the terms by 2: $\sqrt{4+\sqrt{7}}=\sqrt{\frac{8+2\sqrt{7}}{2}}=\frac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}$ Recognizing that $8+2\sqrt{7}$ is $(\sqrt{7}+1{)}^2$ : $\frac{\sqrt{(\sqrt{7}+1{)}^2}}{\sqrt{2}}=\frac{\sqrt{7}+1}{\sqrt{2}}$ Similarly for the second part: $\sqrt{4-\sqrt{7}}=\frac{\sqrt{7}-1}{\sqrt{2}}$ Subtracting the two values: $\frac{\sqrt{7}+1}{\sqrt{2}}-\frac{\sqrt{7}-1}{\sqrt{2}}=\frac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}=\frac{2}{\sqrt{2}}$ Rationalizing the denominator: $\frac{2}{\sqrt{2}}=\sqrt{2}$ Final Answer: $\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\sqrt{2}$ Would you like me to solve a similar radical expression or explain the perfect square transformation in more detail?